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Question Number 59626 by naka3546 last updated on 12/May/19

Rationalize the denominator of (2/(1 − (√(2 + (4)^(1/3) ))))

$${Rationalize}\:\:{the}\:\:{denominator}\:\:{of} \\ $$$$\:\:\:\:\:\:\:\:\frac{\mathrm{2}}{\mathrm{1}\:−\:\sqrt{\mathrm{2}\:+\:\sqrt[{\mathrm{3}}]{\mathrm{4}}}} \\ $$

Answered by MJS last updated on 12/May/19

first use this formula (1/(a−(√b)))=((a+(√b))/((a−(√b))(a+(√b))))=((a+(√b))/(a^2 −b)) then use this formula (1/(a+(b)^(1/3) ))=(((a+(b)^(1/3) (−(1/2)−((√3)/2)i))(a+(b)^(1/3) (−(1/2)+((√3)/2)i)))/((a+(b)^(1/3) )(a+(b)^(1/3) (−(1/2)−((√3)/2)i))(a+(b)^(1/3) (−(1/2)+((√3)/2)i))))= =((a^2 −a(b)^(1/3) +(b^2 )^(1/3) )/(a^3 +b))

$$\mathrm{first}\:\mathrm{use}\:\mathrm{this}\:\mathrm{formula} \\ $$$$\frac{\mathrm{1}}{{a}−\sqrt{{b}}}=\frac{{a}+\sqrt{{b}}}{\left({a}−\sqrt{{b}}\right)\left({a}+\sqrt{{b}}\right)}=\frac{{a}+\sqrt{{b}}}{{a}^{\mathrm{2}} −{b}} \\ $$$$\mathrm{then}\:\mathrm{use}\:\mathrm{this}\:\mathrm{formula} \\ $$$$\frac{\mathrm{1}}{{a}+\sqrt[{\mathrm{3}}]{{b}}}=\frac{\left({a}+\sqrt[{\mathrm{3}}]{{b}}\left(−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\right)\right)\left(\mathrm{a}+\sqrt[{\mathrm{3}}]{{b}}\left(−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\right)\right)}{\left({a}+\sqrt[{\mathrm{3}}]{{b}}\right)\left({a}+\sqrt[{\mathrm{3}}]{{b}}\left(−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\right)\right)\left(\mathrm{a}+\sqrt[{\mathrm{3}}]{{b}}\left(−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\right)\right)}= \\ $$$$=\frac{{a}^{\mathrm{2}} −{a}\sqrt[{\mathrm{3}}]{{b}}+\sqrt[{\mathrm{3}}]{{b}^{\mathrm{2}} }}{{a}^{\mathrm{3}} +{b}} \\ $$

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